Balance the following redox reaction by inserting the appropriate coefficients. H^+ +CrO4^2- +NO 2^- =Cr^3+ +H2O+NO3^-
To balance the given redox reaction:
Step 1: Write down the unbalanced equation:
H^+ + CrO4^2- + NO2^- = Cr^3+ + H2O + NO3^-
Step 2: Separate the equation into half-reactions, one for the reduction and one for the oxidation.
Reduction half-reaction:
CrO4^2- + 4H^+ + 3e^- = Cr^3+ + 2H2O
Oxidation half-reaction:
NO2^- = NO3^-
Step 3: Balance the atoms other than hydrogen and oxygen in each half-reaction.
Reduction half-reaction:
CrO4^2- + 4H^+ + 3e^- = Cr^3+ + 2H2O
Oxidation half-reaction:
3NO2^- = 3NO3^-
Step 4: Balance the oxygen atoms using water molecules.
Reduction half-reaction:
CrO4^2- + 4H^+ + 3e^- = Cr^3+ + 2H2O
Oxidation half-reaction:
3NO2^- + H2O = 3NO3^-
Step 5: Balance the hydrogen atoms using H+ ions.
Reduction half-reaction:
CrO4^2- + 4H^+ + 3e^- = Cr^3+ + 2H2O
Oxidation half-reaction:
3NO2^- + H2O + 2H^+ = 3NO3^-
Step 6: Balance the charges by adding electrons.
Reduction half-reaction:
CrO4^2- + 4H^+ + 3e^- = Cr^3+ + 2H2O
Oxidation half-reaction:
3NO2^- + H2O + 2H^+ + 4e^- = 3NO3^-
Step 7: Multiply each half-reaction by the appropriate coefficient so that the number of electrons is the same in both half-reactions.
Reducing half-reaction:
3CrO4^2- + 12H^+ + 9e^- = 3Cr^3+ + 6H2O
Oxidizing half-reaction:
3NO2^- + H2O + 2H^+ + 4e^- = 3NO3^-
Step 8: Add the two half-reactions together.
3CrO4^2- + 12H^+ + 9e^- + 3NO2^- + H2O + 2H^+ + 4e^- = 3Cr^3+ + 6H2O + 3NO3^-
Simplify:
3CrO4^2- + 15H^+ + 9NO2^- + H2O = 3Cr^3+ + 6H2O + 3NO3^-
Step 9: Cancel out any common species on both sides of the equation.
3CrO4^2- + 15H^+ + 9NO2^- + H2O = 3Cr^3+ + 6H2O + 3NO3^-
(3CrO4^2− + 6H^+ + 6NO2^- = 3Cr^3+ + 6NO3^-)
The balanced equation is:
3CrO4^2- + 6H^+ + 6NO2^- = 3Cr^3+ + 6NO3^-
To balance a redox reaction, we need to make sure that the number of atoms and charge are balanced on both sides of the equation. Here's how you can balance this redox reaction step by step:
Step 1: Write down the unbalanced formula equation:
H^+ + CrO4^2- + NO2^- = Cr^3+ + H2O + NO3^-
Step 2: Identify the atoms undergoing oxidation and reduction:
In this reaction, Cr is being reduced from an oxidation state of +6 to +3, and N is being oxidized from an oxidation state of -3 to +5.
Step 3: Divide the reaction into two half-reactions, one for oxidation and the other for reduction:
Oxidation half-reaction: N2^- → NO3^-
Reduction half-reaction: CrO4^2- → Cr^3+
Step 4: Balance the atoms in each half-reaction:
Oxidation half-reaction: 2 NO2^- → 2 NO3^- (to balance N atoms)
Reduction half-reaction: CrO4^2- → Cr^3+ (1 Cr and 4 O atoms on each side already balanced)
Step 5: Balance the charge in each half-reaction:
Oxidation half-reaction: 2 NO2^- + 4 H^+ → 2 NO3^- + H2O (added 4 H^+ and 2 H2O to balance charge)
Reduction half-reaction: CrO4^2- + 8 H^+ + 6 e^- → Cr^3+ + 4 H2O (added 8 H^+ and 6 e^- to balance charge)
Step 6: Balance the number of electrons transferred between the two half-reactions:
Multiply the oxidation half-reaction by 6 and the reduction half-reaction by 3 to balance the number of transferred electrons:
6(2 NO2^- + 4 H^+ → 2 NO3^- + H2O)
3(CrO4^2- + 8 H^+ + 6 e^- → Cr^3+ + 4 H2O)
This gives us the balanced redox reaction:
6 H^+ + 3 CrO4^2- + 12 NO2^- → 3 Cr^3+ + 6 H2O + 12 NO3^-
So, the balanced coefficients are:
6 H^+ + 3 CrO4^2- + 12 NO2^- → 3 Cr^3+ + 6 H2O + 12 NO3^-
http://www.chemteam.info/Redox/Redox.html
Hint: Cr is +6 on the left and +3 on the right. N is +2 on the left and +5 on the right.